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To be precise, fix $n$, fix a field $k$.

What is the maximal dimension of a subspace of the vector space of all $n\times n$ matrices formed by commutative nilpotent matrices? By commutative I mean all the products of matrices in this subspace are commutative.

(I feel like this can be formulated in terms of Lie algebras, but I don't find a good one. And I think the down-to-earth formulation might make it more accessible.)

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  • $\begingroup$ Commuting matrices can be simultaneously trigonalized over the algebraic closure. If the matrices are nilpotent, their trigonalizations will be strictly upper triangular (because any nonzero entries on the diagonal would survive taking powers, contradicting the nilpotency). So the maximal dimension is $\frac{n\left(n-1\right)}{2}$. $\endgroup$
    – darij grinberg
    Commented Sep 18, 2010 at 22:12
  • $\begingroup$ I get this point, but upper triangular matrices don't necessarily commute with each other, right? $\endgroup$
    – Yuhao Huang
    Commented Sep 18, 2010 at 22:25
  • $\begingroup$ "Commutative" is not a property of a matrix, but of a set of matrices, so I would rather you state the property you want more precisely. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 18, 2010 at 22:28
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    $\begingroup$ I think the substitution of "commuting" for "commutative" would address Qiaochu's concern. $\endgroup$
    – Charles Staats
    Commented Sep 18, 2010 at 22:35
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    $\begingroup$ Related questions: mathoverflow.net/questions/19591/…, mathoverflow.net/questions/19755/…. It came up there that you can achieve the maximal dimension for a commutative subalgebra, 1 plus the floor of n^2/4, by taking scalars plus 2-by-2 block strictly upper triangular matrices. (Apparently the proof of maximality is due to Schur.) Now you can throw out the multiples of the identity to obtain an example of commuting nilpotents having dimension the floor of n^2/4. $\endgroup$
    – Jonas Meyer
    Commented Sep 19, 2010 at 0:28

3 Answers 3

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Try the matrices of the form ${0A\choose 00}$ with A an m by n block (with |m-n|≤1 for maximal dimension).

Spaces of commuting nilpotent matrices have nothing to do with maximal tori or Cartan subalgebras, which consist of semisimple matrices.

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A book "Commutative Matrices", Acad. Press, 1968 by Suprunenko and Tyshkevich (translation from Russian) is devoted, largely, to this question. My cursory look at it tells that the problem seems to be complicated - there are only partial results for small $n$ and particular classes of algebras, expressed in terms of some cumbersome-looking invariants.

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Maybe this paper of Jacobson would help

http://www.ams.org/journals/bull/1944-50-06/S0002-9904-1944-08169-X/S0002-9904-1944-08169-X.pdf

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